Math Problem Statement

Consider the parametric equations below. x = sin2(t), y = sin(2t), 0 ≀ t ≀ πœ‹ 2 Set up an integral that represents the area of the surface obtained by rotating the given curve about the x-axis.

Solution

To set up the integral representing the surface area obtained by rotating the parametric curve given by x=sin⁑2(t)x = \sin^2(t), y=sin⁑(2t)y = \sin(2t) about the xx-axis, we'll use the surface area formula for a parametric curve rotated about the xx-axis:

A=∫ab2Ο€y(dxdt)2+(dydt)2 dtA = \int_{a}^{b} 2\pi y \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt

Given:

  • x=sin⁑2(t)x = \sin^2(t)
  • y=sin⁑(2t)y = \sin(2t)
  • The parameter range is 0≀t≀π20 \leq t \leq \frac{\pi}{2}

Step 1: Compute dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}

  • dxdt=ddt(sin⁑2(t))=2sin⁑(t)cos⁑(t)=sin⁑(2t)\frac{dx}{dt} = \frac{d}{dt} \left( \sin^2(t) \right) = 2\sin(t)\cos(t) = \sin(2t)
  • dydt=ddt(sin⁑(2t))=2cos⁑(2t)\frac{dy}{dt} = \frac{d}{dt} \left( \sin(2t) \right) = 2\cos(2t)

Step 2: Set up the surface area integral

Substitute the values of yy, dxdt\frac{dx}{dt}, and dydt\frac{dy}{dt} into the surface area formula:

A=∫0Ο€22Ο€sin⁑(2t)(sin⁑(2t))2+(2cos⁑(2t))2 dtA = \int_{0}^{\frac{\pi}{2}} 2\pi \sin(2t) \sqrt{\left( \sin(2t) \right)^2 + \left( 2\cos(2t) \right)^2} \, dt

Step 3: Simplify the square root expression

The term inside the square root simplifies as follows:

(sin⁑(2t))2+(2cos⁑(2t))2=sin⁑2(2t)+4cos⁑2(2t)\left( \sin(2t) \right)^2 + \left( 2\cos(2t) \right)^2 = \sin^2(2t) + 4\cos^2(2t)

Thus, the integral becomes:

A=∫0Ο€22Ο€sin⁑(2t)sin⁑2(2t)+4cos⁑2(2t) dtA = \int_{0}^{\frac{\pi}{2}} 2\pi \sin(2t) \sqrt{\sin^2(2t) + 4\cos^2(2t)} \, dt

This is the integral representing the surface area of the curve rotated about the xx-axis.

Would you like me to proceed with further simplification or evaluation of this integral?

Related questions:

  1. What is the physical significance of the surface area obtained by rotating a curve?
  2. How does the parametric representation affect the surface area calculation for rotational surfaces?
  3. Can surface area integrals be evaluated for curves rotated around the yy-axis instead of the xx-axis?
  4. How would you handle the surface area if the limits of the parameter tt were different?
  5. How does the shape of the curve affect the complexity of the surface area integral?

Tip:

When solving problems involving surface areas of rotation, always check if there's symmetry that can simplify the integral before proceeding with full evaluation.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Surface Area of Revolution
Integral Calculus

Formulas

Surface area formula for a parametric curve about the x-axis: A = ∫ 2Ο€y √((dx/dt)^2 + (dy/dt)^2) dt
dx/dt = 2sin(t)cos(t) = sin(2t)
dy/dt = 2cos(2t)

Theorems

Surface Area of Revolution
Chain Rule for Derivatives

Suitable Grade Level

Grades 11-12 or Early College